The answer is -6x + 5y <_ -30
the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5CA%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5CqquadB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquadd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B%5B0-%28-2%29%5D%5E2%2B%5B5-%28-2%29%5D%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B%280%2B2%29%5E2%2B%285%2B2%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B4%2B49%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B53%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5C%5C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-0%29%5E2%2B%281-5%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B3%5E2%2B%28-4%29%5E2%7D)
![\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20BC%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20%5Cboxed%7BBC%3D5%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CC%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2-3%29%5E2%2B%28-2-1%29%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%29%5E2%7D%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B25%2B9%7D%5Cimplies%20%5Cboxed%7BCA%3D%5Csqrt%7B34%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C~%5Chfill%20%5Cstackrel%7BAB%2BBC%2BCA%7D%7B%5Capprox%2018.11%7D~%5Chfill)
Step-by-step explanation:
In the first part, he was travelling with twice the speed, or Vs, for distance
48 - d, for 2 hours.
speed = distance/time
distance = speed * time
First part :
48 - d = 2V * 2
⇒d + 4V = 48
Second part:
d = V * 2
⇒d = 2V
Substitute 2s for d in equation 1.
2s + 4s = 48
6s = 48
s = 8
The speed in the second part is 8 mph.
<u>The speed in the first part is 2s = 2(8) = 16</u>
I hope this helps
The total cost is $66.14. Hope this helps.
<h3>
Short Answer: Yes, the horizontal shift is represented by the vertical asymptote</h3>
A bit of further explanation:
The parent function is y = 1/x which is a hyperbola that has a vertical asymptote overlapping the y axis perfectly. Its vertical asymptote is x = 0 as we cannot divide by zero. If x = 0 then 1/0 is undefined.
Shifting the function h units to the right (h is some positive number), then we end up with 1/(x-h) and we see that x = h leads to the denominator being zero. So the vertical asymptote is x = h
For example, if we shifted the parent function 2 units to the right then we have 1/x turn into 1/(x-2). The vertical asymptote goes from x = 0 to x = 2. This shows how the vertical asymptote is very closely related to the horizontal shifting.
